Number of arrangements in a rectangular table (II) 18 diplomats sit on a rectangular table. Three are from China, four are from Japan, six are from the United States and five are from France. In how many ways can we seat the diplomats at the table so that both the Chinese and the Japanese stay together, but separated from each other?
I thought I had this sorted out, but no.
My proposed solution was the following:
First I allocated the 11 diplomats (US + France) that can sit without restrictions: $$ 11! $$
Then I count the number of places between these diplomats in which the Chinese or the Japanese groups can sit: $$ 11 $$ , since that this is a closed arrangement, the 'last' diplomat is next to the 'first' one.
Then I started by allocating the Chinese group. The Chinese group can stay in 11 places between the $$ 11 $$ diplomats and they can be arranged in $$ 3! $$ ways within themselves: $$ 11 * 3! $$
After the Chinese are seated, we have $$ 10 $$ places left between the diplomats where the Japanese can sit; the Japanese diplomats can be arranged in $$ 4! $$ ways amongst themselves: $$ 10 * 4! $$ .
Lastly, we have to consider the symmetry of the rectangle, meaning that we have been counting these arrangements twice, as the sides of the rectangle are equal two-by-two.
So in my mind we should have in total $$ \frac{11! * 11* 3! * 10 * 4!}{2} = 316 141 056 000 $$ .
However, the number of ways is supposed to be $$ 379 369 267 200 $$ . Can you please help me find what is wrong in my thinking?
Thank you.
 A: I guess I found the supposed solution. Firstly a rectangular table is not a round table. I suppose all diplomats should sit at the same side, so the table is a line. Secondly if we reverse the order we should count it as the same arrangement. (I. e. two arrangements are distinct if at least one diplomat has different set of neighbors.)
Let's consider Chinese delegation as a single metadiplomat and the same for Japanese delegation. Then we have $13$ diplomats, $2$ of which can't sit together. There are $13!$ arrangements in total and $2 \cdot 12!$ of them contain Chinese and Japanese metadiplomats next to each other. So we have $13! - 2 \cdot 12!$ arrangements.
Now remember that Chinese delegation can sit in $3!$ ways and Japanese delegation can sit in $4!$ ways within their places. And reversing order gives the same arrangements. Then we have
$$\frac{(13! - 2 \cdot 12!) \cdot 3! \cdot 4!}{2} = 379369267200$$
arrangements.
P. S. For a round table under the same assumption about distinct arrangements we start with Chinese metadiplomat, have $10$ (of $12$) places for Japanese metadiplomat and $11!$ ways to arrange other diplomats. Taking into account possible reverse of order we get
$$\frac{10 \cdot 11! \cdot 3! \cdot 4!}{2} = 28740096000$$
arrangements, that is $11$ times less than you answer.
If two arrangements are distinct if at least one diplomat has another seat then we should not divide by $2$, but should multiply by $18$ possible options for the leftmost diplomat of Chinese delegation and get
$$10 \cdot 11! \cdot 3! \cdot 4! \cdot 18 = 1034643456000$$
arrangements.
In your reasoning you don't have $11$ places for $11$ diplomats, so you should count only $10!$ cyclic orders, but not $11!$. Then you will get $28740096000$.
A: NEW
I've noticed we could have $J$ at one end, and $C$ at the other, which is also an illegal arrangement.
So I tried this instead, which gives an answer of $465,589,555,200$, but seems logically sound.
Label the chairs $1\dots18$. Note that starting from $1$ is the same as starting from $10$. First place the Chinese trio ($18$ ways times $6$ for internal positions), and then the Japanese ($9\times24$), and then the $11$ remaining diplomats ($11!$), and divide by $2$:
$$\frac{18\cdot6\cdot9\cdot24\cdot11!}{2}=465,589,555,200$$

ORIGINAL POST
I get the correct answer like this:
We have $13$ entities:

*

*6 USA

*5 French

*1 Chinese conglomerate of $3$

*1 Japanese conglomerate of $4$
There are $13!$ ways to arrange these, but then there is the additional rule that we cannot have either $JC$ or $CJ$.
There are $12\cdot11!$ permutations with $JC$ in them, and the same with $CJ$ in them, so the total number of permutations is $13!-2\cdot12!=(13-2)\cdot12!=11\cdot12!=5,269,017,600$.
Next multiply by $3!\cdot4!=144$ for the inner positioning of the Chinese and Japanese, and then a rectangular table has one line of symmetry, so divide by $2$, which gives $379,369,267,200$.
